Suppose we have 1 kg of wood and 1 kg of uranium and if we need to find out how much energy would each of the substance give, we'd have to use Einstein's mass-energy relation as follows:

In the case of wood, $E_{wood} = 1 × (3×10^8)^2 = 9 × 10^{16} J$

In case of Uranium, $E_{uranium} = 1 × (3×10^8)^2 = 9 × 10^{16} J$

Question: According to the equation, both give the same amount of energy. But in reality, 1 kg of uranium will give a lot of energy when compared to 1 kg of wood. What have I missed? Any explanation regarding the $E=mc^2$ problem would be helpful...

2 Answers
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The Einstein's mass-energy relation, $E = mc^2$, gives the total energy content of the system. But this is not the energy we get from the object. When you annihilate an electron with a positron, both particles vanish so that the released energy is equal to the energy of the two particles according to Einstein's formula. But when you burn 1 kg of wood you just transform matter from one form to another (turning organic substances to CO$_2$ and other stuff by oxidation). And when you use nuclear energy of uranium, it just decays into other atoms. Both of these processes release some energy, but it is not equal to $mc^2$, it is much, much less. And because nuclear reactions are much more efficient than burning, you get more energy from 1 kg of uranium than 1 kg of wood.

Particles like neutron, proton and electron have rest mass. The mass-energy equivalence tells that If a particle has rest mass, it has rest energy $m_0c^2$ and may or may not have other kinds of energy such as kinetic & potential energy, etc. This rest energy could be converted to other forms of energy such as heat, light or kinetic energy and the reversal. This means that it is possible to convert an isolated system of particles with mass into a system of particles with less mass or even zero mass (Photon has zero rest mass). Thus, when an electron-positron pair annihilates to form two photons, the energy (kinetic & rest energy) is transferred to the photons that have zero rest energy. Similarly, an energetic photon could create an electron-positron pair on its interaction with strong electric field surrounding a nucleus (which is called pair production).

Off to the topic...

There's something called mass defect ($\Delta m$). When Protons and neutrons combine to form nucleus, the mass that disappears is converted into an equivalent amount of energy called Binding energy.

Assumed mass of nucleons $=Zm_p+Nm_n$. But, this value is greater than the calculated nuclear mass (using mass spectrometers) $m$. Hence, mass defect is $\Delta m=Zm_p+Nm_n-m$. In nuclear physics, it's better to use the term mass-defect energy which is $\Delta mc^2$. In general, the mass defect is calculated by $\Delta m=m_{products}-m_{reactants}$ for nuclear reactions. Then, the equivalent energy is $\Delta mc^2$ (Mass-energy equivalence is applied here). For $U_{235}$, it's found to be $200$ $MeV$ (approx.)

Proving using an example:

Energy per fission of $U_{235} = 200$ $MeV=200×10^6×1.6×10^{-19}J$

No. of atoms in $235$ $g$ of $U_{235}=6.023×10^{23}$ (Avogadro number)